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2x^2+9x-486=0
a = 2; b = 9; c = -486;
Δ = b2-4ac
Δ = 92-4·2·(-486)
Δ = 3969
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{3969}=63$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-63}{2*2}=\frac{-72}{4} =-18 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+63}{2*2}=\frac{54}{4} =13+1/2 $
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